Consider the following equilibrium: 2NH_3(g) N_2 (g) + 3H_2 (g) delta G^0 = 34.

Question

Consider the following equilibrium: 2NH_3(g) N_2 (g) + 3H_2 (g) delta G^0 = 34. kJ Now suppose a reaction vessel is filled with 7.18 atm of ammonia (NH_3) and 8.72 atm of nitrogen (N_2) at 1041. degree C. Answer the following questions a system: under these conditions, will the pressure of N_2 tend to rise or fall? rice fall Is it possible to reverse this tendency by adding H_2? In other words, if you said the pressure of N_2 will tend to rise, can that be changed to a tendency to fall by adding H_2? Similarly, if you said the pressure of N_2 will tend to fall, can that be changed to a tendency to rise by adding H_2? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of H_2 needed to reverse it. Round your answer to 2 significant digits.

Explanation / Answer

dG = 34 kJ = 34000 J/mol

a)

since no H2 is present, NH3 needs to produce more H2 and N2

this means that N2 will increase in its pressure, so answer is "RISE"

b)

Yes, it is possible, if we add enough H2, so we avoid NH3 decomposition

c)

dG = -RT*ln(K)

34000 = -8.314*(1041+273)*ln(K)

K = exp(34000 /( -8.314*(1041+273))) = 0.0445010

then

0.0445010 = N2*H2^3 /(NH3)^2

0.0445010 = 8.72*H2^3 /(7.18)^2

H2 = (0.0445010 *(7.18)^2 / 8.72)^(1/3)

H2 = 0.6407 atm

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